9. The graph of the linear equation \( 2 x+3 y=6 \) cuts the \( y \)-axis at the point: \( \begin{array}{lll}\text { (a) }(2,0) & \text { (b) }(0,3) & \text { (c) }(3,0)\end{array} \)

To find the point where the graph of the linear equation \( 2x + 3y = 6 \) cuts the \( y \)-axis, we need to determine the value of \( y \) when \( x = 0 \). Substituting \( x = 0 \) into the equation: \[ 2(0) + 3y = 6 \] This simplifies to: \[ 3y = 6 \] Now, divide both sides by 3: \[ y = 2 \] Thus, the point where the graph cuts the \( y \)-axis is \( (0, 2) \). However, this point is not listed in the options provided. Let's check the options again: - (a) \( (2, 0) \) - (b) \( (0, 3) \) - (c) \( (3, 0) \) None of these options are correct based on our calculation. The correct point where the graph cuts the \( y \)-axis is \( (0, 2) \).